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Consider a cumulative distribution function $F$.

Take $a: (0,\infty) \rightarrow (0,\infty)$ and $b: (0,\infty) \rightarrow (-\infty,\infty)$. Let $a_s\equiv a(s)$ and $b_s\equiv b(s)$ for all $s \in (0,\infty)$.

Assume $$ [F(a_sx+b_s)]^s=F(x), \quad\forall s>0 $$ could you help me to understand how this implies that

  1. $$ [F(a_{st}x+b_{st})]^s=F\left(\frac{a_{st}}{a_s}x+\frac{b_{st}-b_s}{a_s}\right), \quad \forall s,t>0 $$
  2. $$\left[F\left(\frac{a_{st}}{a_s}x+\frac{b_{st}-b_s}{a_s}\right)\right]^t=F\left(\frac{a_{st}}{a_ta_s}x+\frac{b_{st}-b_s-a_sb_t}{a_sa_t}\right). \quad\forall s,t>0 $$

I don't understand how things are rescaled.

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1 Answer 1

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Because $$ (F(a_s x + b_s))^s = F(x), \quad \forall s > 0, x \in \mathbb{R} $$ then for any $u > 0$, $y \in \mathbb{R}$, take $(s, x) = \left( u, \dfrac{y - b_u}{a_u} \right)$ to get$$ (F(y))^u = F\left( \frac{y - b_u}{a_u} \right). $$ Now, for any $s, t > 0$, $x \in \mathbb{R}$, take $(u, y) = (s, a_{st} x + b_{st})$ to get$$ (F(a_{st} x + b_{st}))^s = F\left( \frac{a_{st} x + b_{st} - b_s}{a_s} \right) = F\left( \frac{a_{st}}{a_s} x + \frac{b_{st} - b_s}{a_s} \right), $$ and take $(u, y) = \left( t, \dfrac{a_{st}}{a_s} x + \dfrac{b_{st} - b_s}{a_s} \right)$ to get\begin{align*} \left( F\left( \frac{a_{st}}{a_s} x + \frac{b_{st} - b_s}{a_s} \right) \right)^t &= F\left( \frac{1}{a_t} \left( \left( \frac{a_{st}}{a_s} x + \frac{b_{st} - b_s}{a_s} \right) - b_t \right) \right)\\ &= F\left( \frac{a_{st}}{a_s a_t} x + \frac{b_{st} - b_s - a_s b_t}{a_s a_t} \right). \end{align*}

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